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          CMU15445 Project
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        <blockquote>
<p>本文章记录<a target="_blank" rel="noopener" href="https://15445.courses.cs.cmu.edu/fall2023/">CMU15445
2023fall</a>课程的实验内容，实验初始源代码<a target="_blank" rel="noopener" href="https://github.com/cmu-db/bustub/archive/refs/tags/v20231227-2023fall.zip">链接</a>，注意不要在github中下载最新的<code>bustub</code>源代码，并采用官方提供的<a target="_blank" rel="noopener" href="https://www.gradescope.com/">Autograde</a>平台进行评测，邀请码KK5DVJ</p>
</blockquote>
<span id="more"></span>
<h1 id="project0-c-primer">Project0 C++ Primer</h1>
<p><a target="_blank" rel="noopener" href="https://postimg.cc/WDqW2K4k"><img src="https://i.postimg.cc/BvMdCf9M/2024-09-10-11-23-39.png" alt="2024-09-10-11-23-39.png"></a></p>
<p>本实验要求为前缀树数据结构完善<code>Get\Put\Remove</code>三种方法，需要注意的是，为了提高前缀树的并发可访问性，我们需要用新的<code>map</code>构建新的节点，以返回新的前缀树<code>Trie</code></p>
<h2 id="task1-写时复制前缀树">Task1 写时复制前缀树</h2>
<h3 id="get"><code>Get</code></h3>
<p>实现<code>Get</code>读取方法需要考虑以下几点</p>
<ul>
<li><p>存在以下两种<strong>边缘情况</strong></p>
<ul>
<li><p><code>Trie</code>为空(<code>root_=nullptr</code>)，<code>default()</code>构造或<code>Remove</code>删除尽有效值节点</p></li>
<li><p><code>key</code>字符串为空，直接查找根节点中的值</p></li>
</ul></li>
<li><p>遍历节点时，以下两种情况直接返回</p>
<ul>
<li>目标<code>key</code>不存在，直接返回<code>nullptr</code></li>
<li>查找到目标节点，并使用<code>std::dynamic_pointer_cast</code>方法成功转化为<code>TrieNodeWithValue</code>进行值读取，需要注意的是<code>std::dynamic_pointer_cast</code>方法在<strong>运行时</strong>检查类型转换</li>
</ul></li>
</ul>
<h3 id="put"><code>Put</code></h3>
<p>关于<code>Put</code>和<code>Remove</code>两种写方法，可以采用递归实现，每一层递归用于向子节点插入<code>key</code>字符串的去头子串，关于该递归方法需要考虑以下几点</p>
<figure class="highlight c++"><table><tbody><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 函数签名</span></span><br><span class="line"><span class="keyword">template</span> &lt;<span class="keyword">class</span> <span class="title class_">T</span>&gt;</span><br><span class="line"><span class="function"><span class="type">static</span> <span class="keyword">auto</span> <span class="title">RecursivePut</span><span class="params">(<span class="type">const</span> std::shared_ptr&lt;<span class="type">const</span> TrieNode&gt; &amp;node, std::string_view key, T value)</span></span></span><br><span class="line"><span class="function">    -&gt; std::shared_ptr&lt;<span class="type">const</span> TrieNode&gt; </span>{</span><br></pre></td></tr></tbody></table></figure>
<ul>
<li><p>两种边缘情况</p>
<ul>
<li><code>key</code>字符串为空，直接将值插入当前节点，但需要注意继承原先节点的<code>children_</code>子节点映射</li>
<li><code>key</code>字符串长度为1，考虑两种情况
<ul>
<li>该字符已存在对应子节点，需要继承原节点的<code>children_</code>映射并更新<code>value_</code></li>
<li>该字符不存在对应子节点，则用<code>value_</code>构造新的子节点，并与父节点连接</li>
</ul></li>
</ul></li>
<li><p>递归插入，同样考虑两种情况</p>
<ul>
<li><p>当前字符串<code>key</code>的首字符已存在子节点，则需要向递归函数中传入已存在的子节点指针</p></li>
<li><p>当前字符串首字符不存在子节点，则传入新构造的空子节点指针</p>
<blockquote>
<p>此处存在缺陷是，函数传入的空子节点，进入函数体内部后又会被重新<code>Clone()</code>一次，舍弃原先的空节点，造成无用的重复构造</p>
</blockquote></li>
</ul></li>
</ul>
<h3 id="remove"><code>Remove</code></h3>
<figure class="highlight c++"><table><tbody><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 函数签名</span></span><br><span class="line"><span class="function"><span class="type">static</span> <span class="keyword">auto</span> <span class="title">RecursiveRemove</span><span class="params">(std::shared_ptr&lt;<span class="type">const</span> TrieNode&gt; node, std::string_view key)</span></span></span><br><span class="line"><span class="function">    -&gt; std::shared_ptr&lt;<span class="type">const</span> TrieNode&gt; </span></span><br></pre></td></tr></tbody></table></figure>
<ul>
<li>三种边缘情况
<ul>
<li><code>key</code>字符串为空，需要删除根节点中的值</li>
<li><code>key</code>字符串的首字符不存在，说明待删除的键不存在，直接返回原节点</li>
<li><code>key</code>字符串长度为1
<ul>
<li>该字符对应的节点不存在值，说明该键值对不存在，直接返回原节点</li>
<li>键值对存在，但是值节点不是叶子节点，需要将值节点降级为内部节点</li>
<li>键值对存在，且值节点是叶子节点，则需要<code>children_.erase()</code>删除节点</li>
</ul></li>
</ul></li>
<li>递归删除，考虑两种情况
<ul>
<li>返回的子节点既不是值节点，<code>children_</code>也为空，则需要摒弃该子节点</li>
<li>否则，将新的子节点连接到当前节点</li>
</ul></li>
</ul>
<h2 id="task2-并发键值存储">Task2 并发键值存储</h2>
<p>在前缀树的并发访问中，题目要求读之间兼容、写操作互斥、写操作进行时，读操作能够从旧根中开始。该并发设计最难在于最后一点，但是<code>Task1</code>的正确实现能够满足这一点</p>
<p><img src="https://15445.courses.cs.cmu.edu/fall2023/project0/trie-02.svg" style="zoom: 25%;"></p>
<p>如上图所示，当读写操作并行时，只要读操作能够获取原先的旧根指针，就能在<code>Get</code>方法的逻辑中正确读下去，原先逻辑删除的节点仍然物理存在</p>
<h1 id="project1-buffer-pool">Project1 Buffer Pool</h1>
<h2 id="task1-lrukreplacer">Task1 LRUKReplacer</h2>
<p>​
在测试<code>LRUKReplacer</code>类型中<code>LRU-K</code>算法时，重点在于<code>Evict()</code>驱逐方法的实现。</p>
<p>​
关于<code>LRU</code>算法，我们只关注缓冲池中每一个页帧最近一次访问的时间，所谓<code>least recently used</code>算法，即<code>LRU-1</code>算法，最近<strong>一次</strong>使用距离现在最远的页帧是我们需要淘汰的对象，但<code>LRU-1</code>不能体现<code>LRU</code>算法<strong>最近最少</strong>使用的思想，因为页帧的淘汰都仅考虑最近一次的访问时间戳（而不是最近多次），其存在弊端就是未能考虑到页帧访问的频率：假设页帧2的访问频率远低于页帧1，而页帧2和页帧1同时存在于缓冲池中，但是页帧2最近一次访问时间近于页帧1，那么在驱逐时可能会将访问频率高得多的页帧1驱逐出去。</p>
<p>​
因此，在<code>LRU-K</code>算法中，我们跟踪了缓冲池中每一个页帧最近<code>K</code>次访问的时间戳，每当要进行淘汰时，我们驱逐倒数第<code>K</code>次访问距离现在最远的页帧，在这样巧妙的比较下，该算法隐式地将页帧访问频率考虑了进去。</p>
<p>​
参考以下图片，可以看到容量为2的缓冲池中缓存了A，B的访问频率远小于A，但是B最近的访问近于A，在<code>LRU</code>算法中，此时我们淘汰A；但是在<code>LRU-5</code>算法中，我们将当前时间戳和页帧倒数第5次访问的时间戳做差，发现B的差值更大，说明B的访问频率低于A，淘汰对象为B</p>
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